Arrange the energy of the 2s orbital in the f | vedclass

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Question

(C) The energy of an orbital in a multi-electron atom depends on the effective nuclear charge $(Z_{eff})$ and the principal quantum number $(n)$.For a

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$E_{2s(H)} > E_{2s(Li)} > E_{2s(Na)} > E_{2s(K)}$

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(C) The energy of an orbital in a multi-electron atom depends on the effective nuclear charge $(Z_{eff})$ and the principal quantum number $(n)$.For a given orbital like $2s$,as the atomic number $(Z)$ increases,the effective nuclear charge experienced by the electron increases due to the shielding effect of inner electrons.However,in hydrogen $(H)$,there is only one electron and no shielding,making the $2s$ orbital energy higher compared to multi-electron atoms where the electron is more tightly bound to the nucleus.As we move down the group $IA$ ($Li$ to $K$),the atomic size increases and the screening effect increases,which stabilizes the $2s$ orbital,thereby decreasing its energy.Therefore,the correct decreasing order of energy is $E_{2s(H)} > E_{2s(Li)} > E_{2s(Na)} > E_{2s(K)}$.

Arrange the energy of the $2s$ orbital in the following atoms in decreasing order: $H, Li, Na, K$.

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